A mass m performs oscillations of period $T$ when hanged by spring of force constant $K$. If spring is cut in two parts and arranged in parallel and same mass is oscillated by them, then the new time period will be
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Two simple pendulums whose lengths are $100 cm$ and $121 cm$ are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum, will the two be in phase again
A potential is given by $V(x)=k(x+a)^2 / 2$ for $x < 0$ and $V(x)=k(x-a)^2 / 2$ for $x > 0$. The schematic variation of oscillation period $T$ for a particle performing periodic motion in this potential as a function of its energy $E$ is
Two bodies performing $SHM$ have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is
A particle executes $SHM$ with amplitude of $20 \,cm$ and time period is $12\, sec$. What is the minimum time required for it to move between two points $10\, cm$ on either side of the mean position ..... $\sec$ ?
Time period of a simple pendulum is $T$ inside a lift when the lift is stationary. If the lift moves upwards with an acceleration $g / 2,$ the time period of pendulum will be
When a body of mass $1.0\, kg$ is suspended from a certain light spring hanging vertically, its length increases by $5\, cm$. By suspending $2.0\, kg$ block to the spring and if the block is pulled through $10\, cm$ and released the maximum velocity in it in $m/s$ is : (Acceleration due to gravity $ = 10\,m/{s^2})$